Specific heat, the calorie, Joule and heat transfer
  

Dr. Walt Volland revised June 28, 2013

Energy transferred from a hot object to a cold object is called heat.  This how hot and cold are defined. Heat moves from high temperature hot objects to lower temperature cold objects. When you touch something that feels hot to you heat (energy) is moving to you from the hot object. When you touch an object that feels cold like an ice cube heat moves from you into the ice cube. You are hotter at 98.6oF ( 37oC) and losing energy to the ice cube at 32oF (0oC ).

Energy is labeled with many unit types that depend on the measurement system.
The SI energy unit is the joule.  4.186 Joule = 1 calories.  Another common energy unit is the calorie. 

The calorie, cal, is defined as the amount of energy (heat) needed to increase the temperature of one gram of water by 1oC.  The kilocalorie, kcal, equals 1000 small calories.  To make life more complicated Nutritional energy values are reported in dietetic calories or the big Calorie. The nutrition Calorie is really equal to the kilocalorie or 1000 cal.   1 C  = 1000 calories = 1 kcal.

The little calorie is so small that the calorie yield from a slice of buttermilk bread equals 80 C or 80,000 calories. This link provides Calorie values for thousands of food items.
http://www.calorieking.com/foods/

Specific heat is a physical property of materials. It is used to describe heat transfer. The specific heat of a material is an intensive property that is independent of the amount of a substance. It is defined as the amount of energy needed to heat one gram of the substance by 1 degree Celsius. The mathematical expression is    Specific heat =   calories  /  grams oC

Specific heat is essential when designing engine cooling systems, constructing electronic circuits so they do not overheat, designing refrigeration equipment and everything that involves heat transfer.

The amount of heat needed to increase the temperature of any object can be figured using this relationship.
Heat in calories = mass in gram x temperature change oC  x specific heat [ cal / gram oC]
The formula has four terms in it. You can calculate any of them if you have the other three.
Example:
How many calories are needed to heat a cup of 200 grams of water from room temperature at 20oC to 100oC? The specific heat for water is 1 calorie / g oC.
Answer:
Determine what you are supposed to figure. The question asks for the “heat”.
Identify the quantities given in the question.
Mass = 200 grams
Temperature change in degrees
oC   =  Tfinal  -  T initial  =   100oC  - 20oC  = 80oC
Specific heat  = 1 cal / g oC

Heat = ( 200 g ) ( 80oC ) ( 1 cal / g oC)   =  16000 calories
The grams and degrees C cancel.